I asked a similar question here more than two years ago. I did not get an answer to my complete satisfaction. I would like to reiterate the problem again.

The local electric field of a monochromatic radiation is nonzero and varies sinusoidally in a predictable fashion. For example, the electric field of an ideal monochromatic radiation is described by $${\bf E}({\bf r},t)={E}_0\hat{{ \varepsilon}}\cos({\bf k}\cdot{\bf r}-\omega t),$$ at any location ${\bf r}$, is nonzero at any time $t$ and varies with time in a predictable manner. Here, $E_0$ is a fixed number and so is $\omega$ (the frequency of the radiation), and $\hat{\varepsilon}$ represents the constant polarization vector.

In contrast, assuming that the electric field at any location of a Blackbody cavity is due to an incoherent superposition of electric fields of all frequencies, polarizations (and all amplitudes?), can we rigorously come up with a mathematical expression for the local electric field at any time $t$ for the blackbody radiation? I am interested in getting a mathematical formula that properly represents the local electric field of incoherent thermal radiation.

  • $\begingroup$ Would an expression for the expectation value of $\mathbf{E}(\mathbf{r})$ be sufficient? I doubt that you can get the exact expression for $\mathbf{E}(\mathbf{r})$ itself; we never really know the microstate of a system in thermal equilibrium (if we did, we wouldn't be using thermodynamics to describe it.) $\endgroup$ Aug 31, 2020 at 13:52
  • $\begingroup$ Do you mean a thermal average? I think that will do. $\endgroup$
    – SRS
    Aug 31, 2020 at 14:58

1 Answer 1


Using Cartesian coordinates you could write something like: \begin{eqnarray} \vec{E}(\vec{r},t) & = & \int E_{0}(\omega) \left(\cos (\omega[t - x/c]) \hat{j} + \cos(\omega[t -x/c] +\phi_{x,\omega}) \hat{k}\right)\ d\omega \nonumber \\ & + & \int E_0(\omega)\left( \cos (\omega[t - y/c]) \hat{i} + \cos(\omega[t -y/c] +\phi_{y,\omega}) \hat{k}\right)\ d\omega \nonumber \\ & + & \int E_0(\omega)\left( \cos (\omega[t - z/c]) \hat{i} + \cos(\omega[t -z/c] +\phi_{z, \omega}) \hat{j}\right)\ d\omega . \nonumber \end{eqnarray} This represents the sum of three unpolarised beams of light travelling in three directions, each with equal power, summed over all frequencies. In each term, the $\phi_{i,\omega}$ term represents a random phase between $0$ and $2\pi$.

The time averaged value of $\vec{E}\cdot \vec{E}$ is $$\left< \vec{E}\cdot \vec{E}\right> = 3 \int E_0^2(\omega)\ d\omega $$ and so the time-averaged energy density of the electromagnetic fields would be $$ 6\epsilon_0 \int E_0^2(\omega)\ d\omega = \int \frac{\hbar \omega^3}{\pi^2 c^4} \left(\frac{1}{\exp[\hbar \omega/k_BT]-1}\right)\ d\omega\ , $$ where the term on the right is the total energy density of a blackbody radiation field and the extra factor of 2 on the left hand side accounts for equal energy density in the magnetic field. Thus $$E_0(\omega) = \left[ \frac{\hbar \omega^3}{6\epsilon_0\pi^2 c^4} \left(\frac{1}{\exp[\hbar \omega/k_BT]-1}\right) \right]^{1/2} $$

Of course, the time-average of the electric field itself is zero.


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